Re: [Haskell-cafe] groupBy without Ord?

30 Mar 2014


      Am 03/30/2014 06:00 PM, schrieb Kim-Ee Yeoh:
...
On Sun, Mar 30, 2014 at 9:06 PM, martin mailto:martin.drautzburg@web.de> wrote:
groupEq (x:xs) = (groupEq a) ++ [x] ++ (groupEq b)
            where
                a = filter (==x) xs
                b = filter (/= x) xs
Why the need to recurse on "a"? It's a list of identical elements!
Silly me. I had just mechanically modified the textbook quicksort.